Kepler’s Validation of G

Kepler’s Validation of G

Cavendish discovered the universal gravitational constant G in his
laboratory in 1798 with a value of 6.74E-11. Today the recognized
value is 6.67259E-11. The formula P=sqrt((4pi2r3)/(G(msun+mplanet)),
which was derived from Kepler’s planetary laws, provides a link
between G and the orbital periods of free fall bodies in the solar
system. If we apply this formula with an adjusted value of
G=6.66662E-11 (for a better fit) we get the following differences in
days between observed (in Pluto’s case predicted) and calculated
orbital periods of the selected planets and moons:

OBS DAYS CALC DAYS DAYS DIFF
MERCURY 87.96 87.97 0.0087428
VENUS 224.68 224.70 0.0177057
EARTH 365.26 365.25 -0.0061423
MARS 686.98 686.87 -0.1145676
JUPITER 4332.71 4334.48 1.7681635
SATURN 10759.10 10759.00 -0.0982464
URANUS 30707.41 30707.07 -0.3376758
NEPTUNE 60198.50 60198.51 0.0112854
PLUTO 90474.90 90763.22 288.3172342

MOON 27.32 27.27 -0.0513968

IO 1.77 1.77 0.0000000
EUROPA 3.55 3.55 0.0000000
GANYMEDE 7.15 7.15 0.0000000
CALLISTO 16.69 16.69 0.0000000

The results from the formula derived from Kepler’s planetary laws with
the adjusted value of G validate the constant G as a universal
constant in laboratory, heliocentric, geocentric and joviancentric
environments. In all cases except Pluto’s and Jupiter’s, the
difference is less than +-0.4 days. The large difference for Pluto may
indicate an error in the predicted orbital period or the mean distance
from the sun or the assumed mass. For Jupiter it is remarkable that
while the planet has a difference of 1.7 days in 4334 days, his four
moons have no difference at all.

Peter Riedt

On Mar 15, 10:32*am, Peter Riedt wrote:
Kepler’s Validation of G

Cavendish discovered the universal gravitational constant G in his
laboratory in 1798 with a value of 6.74E-11. Today the recognized
value is 6.67259E-11. The formula P=sqrt((4pi2r3)/(G(msun+mplanet)),
which was derived from Kepler’s planetary laws, provides a link
between G and the orbital periods of free fall bodies in the solar
system. If we apply this formula with an adjusted value of
G=6.66662E-11 (for a better fit) we get the following differences in
days between observed (in Pluto’s case predicted) and calculated
orbital periods of the selected planets and moons:

* * * * * * * * OBS DAYS * * * *CALC DAYS * * * DAYS DIFF
MERCURY * * * * 87.96 * 87.97 * 0.0087428
VENUS * * * * * 224.68 *224.70 *0.0177057
EARTH * * * * * 365.26 *365.25 *-0.0061423
MARS * * * * * *686.98 *686.87 *-0.1145676
JUPITER * * * * 4332.71 4334.48 1.7681635
SATURN * * * * *10759.10 * * * *10759.00 * * * *-0.0982464
URANUS * * * * *30707.41 * * * *30707.07 * * * *-0.3376758
NEPTUNE * * * * 60198.50 * * * *60198.51 * * * *0..0112854
PLUTO * * * * * 90474.90 * * * *90763.22 * * * *288.3172342

MOON * * * * * *27.32 * 27.27 * -0.0513968

IO * * * * * * *1.77 * *1.77 * *0.0000000
EUROPA * * * * *3.55 * *3.55 * *0.0000000
GANYMEDE * * * * * * * *7.15 * *7.15 * *0.0000000
CALLISTO * * * * * * * *16.69 * 16.69 * 0.0000000

The results from the formula derived from Kepler’s planetary laws with
the adjusted value of G validate the constant G as a universal
constant in laboratory, heliocentric, geocentric and joviancentric
environments. In all cases except Pluto’s and Jupiter’s, the
difference is less than +-0.4 days. The large difference for Pluto may
indicate an error in the predicted orbital period or the mean distance
from the sun or the assumed mass. For Jupiter it is remarkable that
while the planet has a difference of 1.7 days in 4334 days, his four
moons have no difference at all.

Peter Riedt

Correction, the adjusted value of G should be 6.6662E-11 not
6.66662E-11.